Global Population & Energy Systems: Dynamics and Applications

Posted on Aug 3, 2025 in Geology

Population Growth Dynamics

Current World Population: 7.3 Billion

World population today: 7.3 Billion

The Population Hourglass

7 billion now (2011)
140 million born each year
57 million die each year
108 billion estimated to have ever lived on Earth

Population Density: Understanding Distribution

Population density (number of persons per square km) is more useful than its inverse. The inverse of population density is the area per person.

Key Factors Affecting Population

Population
Growth rate
Life expectancy
Fertility
Developing Countries: High growth rate, lower life expectancy

From Poverty to Prosperity: A Historical Shift

Two centuries ago, most people were poor and had short lives. The Industrial Revolution led to increased wealth and longer lifespans. This trend continues today.

Compound Interest and Population Growth

Compound monthly: amount A at the end of the year is given by A = (1+r/12)^12

Compound daily: A = (1+r/365)^365

Compounded n times annually: A = (1 + r/n)^n

The amount after t years is given by A(t) = e^(rt)

The exponential growth rate (e^rt) increases faster than a linear rate (rt).

Understanding Exponential Growth

Exponential growth, characterized by increasingly faster growth (a “boom”), is inherently unsustainable.

The Rule of 70: Doubling Time

The sum S_k after doubling (k-1) times = 2^k – 1. When (k-1) is sufficiently large, S_k ≈ 2^k.

Energy Generation and Consumption

Historical Energy Sources: Muscle Power

Muscle power: animal and slave muscles
Horsepower (circa 1900)

The Rise of Mechanical and Electrical Power

The invention of machines, fueled by burning fossil fuels, led to the development of mechanical and electrical power.

AC Electricity (1888, 1896)

Traditional vs. Modern Energy Systems

Traditional: Human and animal power, wood; thinly and widely distributed in small quantities; lower efficient utilization.
Modern: Coal, oil, gas, hydro, and nuclear power; large-scale, localized, and mass production; efficient, affordable, and reliable; utilizes machines to convert energy into mechanical energy, heat, and electricity.

Two Major Energy Transitions

Post-Industrial Revolution: Steam engines powered by coal, diversification of energy end-use technologies, and new energy supply sources like electricity and internal combustion engines.
Oil has since overtaken coal as the most dominant energy source.

Steam Engines

Steam engines convert the chemical energy from burning coal into mechanical energy.

Internal Combustion Engines & Steam Turbines

Otto Petrol Engine: Primarily used in road vehicles.
Diesel Engine: Powers LNG tankers, bulk cargo vessels, and massive container ships.
Jet Engine: Essential for aircraft.

Measuring Energy

Energy is measured in Joules (J) and kilowatt-hours (kWh).

Power (P) is the rate of energy transfer (E/T), measured in Watts (W).
1 Manpower = 75 W (or 75 J/s) or 0.1 horsepower (hp).

Fossil Fuels: Primary Energy Resources

Fossil fuels are essential for building and operating machines.

Impact of Fossil Fuels: Pros and Cons

Pros: Cheap energy from fossil fuels led to the invention of machines.
Pros: Increased food availability and improved hygiene contributed to population growth.
Cons: Explosive growth in energy consumption and population has a significant impact on the environment.

The IPAT Equation: Environmental Impact

Impact (I) = Population (P) × Affluence (A) × Technology (T)

Affluence (A): Represents wealth, often measured by Gross Domestic Product (GDP).

Significant disparity exists in per capita energy use between industrialized and developing countries, rendering the world average less meaningful.

Applications of Fossil Fuels

Fossil fuels offer high energy density, satisfying energy demands for industrial production, heating, electricity, and transportation. They are also crucial for manufacturing commercial and industrial products.

Combustion of Fossil Fuels

The burning of fossil fuels involves the oxidation of hydrocarbons.

Coal/Oil/Gas + Oxygen → Heat + Water + Carbon Dioxide
C_xH_y + (x+y/4)O₂ → Heat + (y/2)H₂O + xCO₂

This process leads to air pollution and contributes to global warming.

Characteristics of Fossil Fuels

Primary Uses: Electricity (coal), transportation (oil), food production (irrigation, fertilizers, pesticides), and chemicals (coal, oil, natural gas).
Non-renewable, main energy source (80%).
Advantages: Cheap, abundant, high energy density, easy to store and transport (except natural gas), and relatively safe.

Coal: Formation and Types

Coal is the most abundant fossil fuel, formed from dead plants covered by layers of water and dirt. Heat and pressure from overlying layers transform the plant remains, trapping their energy.

Lignite: Immature, ~30% Carbon, 5k-7kBtu/lb
Sub-bituminous: ~40% Carbon, 8k-10k Btu/lb
Bituminous: 50-70% Carbon, 11k-15k Btu/lb
Anthracite: Mature, ~90% Carbon, ~14k Btu/lb

Coal Mining Methods

Surface Mining: Used when coal is buried less than approximately 60 meters. Involves removing topsoil and rock layers to expose coal. After extraction, dirt and rock are returned, and the area is replanted.
Underground Mining (Deep Mining): For coal buried 300 meters or more below the surface. Miners use elevators to descend shafts and operate machines to extract coal.

Coal Processing

After extraction, coal is conveyed to a preparation plant to remove rocks, dirt, sulfur, and other unwanted materials, thereby increasing its heating value.

Coal Transportation

Coal is transported via various methods: shipping (often more costly than mining), trucks and conveyors (for nearby sites), trains, barges, ships, and even pipelines (for crushed coal mixed with water).

Coal Mining Safety Risks

Risks include suffocation, gas poisoning, roof collapse, and gas explosions.

Uses of Coal

Coal is used for approximately 41% of electricity generation, and in the production of steel, cement, plastics, tar, synthetic fibers, and medicines.

Coal Energy Density and Power Generation

Energy Density: 24 MJ/kg or 6.67 kWh/kg.
Coal power plants typically obtain about 2.0 kWh/kg by burning coal.

Environmental Impact of Coal

Emissions include CO2, particulate matter (soot or fly ash containing SiO2, Al2O3, Fe2O3, and oxides of Ca, Mg, K), mercury, radioactive uranium, SOx (0.5-2 kg/GJ), NOx (0.1-1 kg/GJ), and volatile organic compounds. Notably, coal ash can be more radioactive than nuclear waste.

Acid Rain: Rain becomes acidic when it mixes with atmospheric oxides like CO2, nitrogen oxides, and sulfur dioxide. This has non-local and adverse effects.

Petroleum and Natural Gas

Petroleum and natural gas form from dead sea plants and animals, buried on the ocean floor and covered by sand.

Formation of Petroleum and Natural Gas

1. Generation: Accumulated organic debris from microorganisms is buried in fine-grained sedimentary source rock to form kerogen. Subjected to geothermal heating at appropriate depths (50-200°C), kerogen cracks to form hydrocarbons. The oil window is 50-150°C, and the gas window is 150-200°C.
2. Migration: Hydrocarbons migrate from the source rock along faults to reservoir rock or the ground surface.
3. Entrapment: Hydrocarbons accumulate and are stored in porous reservoir rock (sandstone or limestone), protected by an impermeable seal rock, typically with gas on top of oil, on top of water.

Petroleum Production Methods

Exploration Drilling: Extensive drilling determines the size and boundaries of oil fields and estimates recoverable reserves.
Primary Recovery: Pressurized oil rises naturally up the borehole. This method can recover up to 30% of the oil in the reservoir rock’s fine pores.
Secondary Recovery: Once natural pressure dissipates, oil flow is forced out by injecting gas into the gas cap above the oil layer or water into the aquifer below. This can recover up to 65% of the oil.

Oil Refining and Products

Hydrocarbon products are obtained from crude oil through fractional distillation, with approximately 60% used for transport.

A barrel of crude oil equals 42 US gallons. Distillation can increase the volume of refined products to over 44 gallons.

The Role of OPEC

The Organization of the Petroleum Exporting Countries (OPEC), a permanent intergovernmental organization, was founded in 1960 by Iran, Iraq, Kuwait, Saudi Arabia, and Venezuela. OPEC coordinates oil production policies to stabilize the oil market, ensure reasonable returns for producers, and maintain stable oil supplies for consumers.

Hubbert’s Peak Theory

Hubbert’s Peak theory posits that the rate of petroleum production follows a bell-shaped curve. The peak is reached when approximately half of the total oil reserve has been extracted.

M. King Hubbert’s US Oil Reserve Prediction

Total oil discovered (Q_d)
Total oil produced (Q_p)
Oil remaining (Q_r) = Q_d – Q_p

Hubbert’s prediction began to fail after 1995, as it did not account for potential resource growth, new technologies, commercial factors, or geopolitical impacts on production.

Natural Gas: Properties and Value

Natural gas has a higher energy value than coal or crude oil.

Coal: 24 MJ/kg
Crude Oil: 41 MJ/kg
Natural Gas: 44 MJ/kg

Natural Gas Purification

Extracted natural gas is transported via gathering pipelines (small-diameter, low-pressure pipes). Purification involves four main processes to remove impurities:

Oil and Condensate Removal
Water Removal
Separation of Natural Gas Liquids
Sulfur and Carbon Dioxide Removal

Natural Gas Transportation

Pipelines: Used for onshore transport.
LNG Carriers: Liquefied Natural Gas (LNG) is transported across seas in specialized carriers (120,000 m³ to 140,000 m³ capacity). A typical 20-day voyage can result in 2-6% volume loss.

Natural Gas Storage

Natural gas is typically stored in underground reservoirs.

Hydraulic Fracturing (Fracking)

Most oil and gas deposits are found in porous sandstones and coarse-grained limestones, which act like hard, non-compressible sponges, saturated with water, oil, or gas in pores ranging from tens to hundreds of micrometers. Shale rocks, however, are mudstones with nanoscale pores, trapping gas or oil. Hydraulic fracturing has enabled the economical recovery of natural gas and oil from these shale and limestone formations.

Combustion and Oxidation Processes

High Temperature (Combustion): Hydrocarbons (C_nH_2n+2, C_nH_2n)

CH₄ (methane) + 2O₂ → CO₂ + 2H₂O + Energy

Low Temperature (Metabolism): Carbohydrates (C_n(H_2mO_m)) in the presence of catalysts.

C₆H₁₂O₆ (glucose) + 6O₂ → 6CO₂ + 6H₂O + Heat

Approximately 35% of CO2 emissions come from burning coal, and 60% from burning other hydrocarbons.

Thermodynamics of Heat Engines

Heat is most readily obtained through combustion.

Burning: Oxidation of a solid in the atmosphere with oxygen.
Boiling: Changing a liquid to a gas, typically at high temperatures.

Phase Transitions:
Gas to liquid: Condensation
Gas to solid: Deposition
Liquid to gas: Vaporization
Liquid to solid: Freezing
Solid to gas: Sublimation
Solid to liquid: Fusion

Heat energy spontaneously flows from regions of high temperature to regions of low temperature. The amount of mechanical work or energy produced from heat by an engine is governed by the Laws of Thermodynamics.

What is Thermodynamics?

Thermodynamics is the study of heat energy conversion into other forms of energy, subject to physical constraints.

The First Law of Thermodynamics

The First Law states that it is impossible to obtain more energy than what flows through the engine. It is essentially the law of conservation of energy.

ΔU (change in internal energy) = Q (heat added) – W (work done by system)

Internal energy (U) is related to the kinetic energy of molecules; it represents potential to do work even when no work is being done.

The Second Law of Thermodynamics

The Second Law states that it is impossible to convert all heat energy flowing through an engine into work; only a portion can be converted. This law dictates the direction of energy transfer.

Reversible Process: A process that can be reversed by an infinitesimally small change in applied conditions.
Irreversible Process: Most processes observed in daily life are irreversible.

When two systems at different temperatures are in thermal contact, heat flows spontaneously from high to low temperature; the reverse process is not possible. This law expresses the degradation of energy.

The Concept of Entropy (S)

Boltzmann’s entropy equation: S = k lnΩ

Where k is the Boltzmann constant (k = 1.38 × 10^-23 J/K) and Ω is the multiplicity of the macrostate.

S_total = k ln(Ω₁ × Ω₂) = k ln(Ω₁) + k ln(Ω₂) = S₁ + S₂

For an isolated system, disorder (entropy) generally increases.

Kinetic Theory of Ideal Gases

Heat Transfer in an Ideal Gas Cylinder

A piston and the far end of the cylinder are in thermal contact with heat reservoirs. A heat reservoir is a large energy source where a finite heat transfer (Q) does not change its temperature. ΔU = U_f – U_i = Q – W.

Work Done by Gas: Area Under the Curve

The work done by the gas is represented by the area under the curve from initial (i) to final (f) states. Expansion results in positive work; compression results in negative work. Work done by the surroundings is opposite to work done by the gas.

Types of Thermodynamic Processes

Isobaric Process (Constant Pressure): W = PΔV, ΔU = Q – W
Isothermal Process (Constant Temperature): ΔU = 0, so Q = W. Energy entering as heat must leave as work. W = ∫ P dV = ∫ nRT (dV/V), W = nRT ln(V_f / V_i)
Isochoric Process (Constant Volume): W = 0, ΔU = Q
Cyclical Process: ΔU = 0, so Q = W
Adiabatic Process (No Heat Exchange): Q = 0, ΔU = -W.

Achieved by thermally insulating the system walls or by processes occurring too quickly for heat exchange. For an ideal gas, ΔU = nC_vΔT, where C_v is the specific heat capacity at constant volume. Rapid gas expansion (Q=0) leads to increased volume and decreased pressure and temperature. PV^γ = constant and TV^γ-1 = constant. P_f(isothermal) = P_i(V_i/V_f), P_f(adiabatic) = P_i(V_i/V_f)^γ. Since (V_i/V_f) < 1 and γ > 1, P_f(adiabatic) < P_f(isothermal).

Molar Specific Heat

Constant Volume: Q = nC_vΔT (W=0, ΔU=Q)
Constant Pressure: Q = nC_pΔT, where C_p = C_v + R

R = 8.315 J/mole/K (ideal gas constant)

For given n and ΔT, Q_p > Q_v, as Q_p accounts for both internal energy increase and work done by the system.

The ratio of molar specific heats (γ = C_p/C_v) is approximately 1.67 for monatomic, 1.40 for diatomic, and 1.30 for polyatomic gases.

Heat Engines

A heat engine absorbs heat (Q_h) from a high-temperature (T_h) hot reservoir, converts part of it into work (W) in a cyclic process, and expels the remainder to a cold reservoir (T_c).

Q_h = Q_c + W

The thermal efficiency (e) in one cycle is: e = W/Q_h = (Q_h – Q_c) / Q_h = 1 – Q_c/Q_h < 1

Efficiency (e) can only equal 1 if Q_c = 0, meaning all heat from the hot reservoir is converted to work (Q_h = W, e=100%). This “perfect” scenario is impossible due to the Second Law of Thermodynamics.

Heat Engine Efficiency

Q_c / T_c ≤ Q_h / T_h, or Q_c / Q_h ≤ T_c / T_h

e = 1 – Q_c / Q_h ≤ 1 – T_c / T_h

(A smaller T_c/T_h ratio results in higher efficiency.)

The First Law dictates that e cannot be greater than 1; the Second Law dictates that e cannot be equal to 1.

Heat Pumps

A heat pump operates as the reverse of a heat engine. Energy (Q_c) is extracted from a cold reservoir and transferred to a hot reservoir (Q_h). Since this is not a natural direction of energy transfer, work (W) must be input into the system.

Coefficient of Performance (COP)

Cooling Mode: COP = Q_c / W ≤ T_c / (T_h – T_c)
Heating Mode: COP = Q_h / W ≤ T_h / (T_h – T_c)

(Temperatures T are in Kelvin.)

COP improves as the temperature difference (ΔT = T_h – T_c) decreases.

The Carnot Engine

Isothermal Expansion (A → B): At T_h, the gas absorbs heat |Q_h| and does work W_ab by raising the piston.
Adiabatic Expansion (B → C): With thermally non-conducting walls, no heat enters or leaves. Temperature falls from T_h to T_c. Work W_bc is done.
Isothermal Compression (C → D): At T_c, the gas expels energy Q_c. Work W_cd is done.
Adiabatic Compression (D → A): With non-conducting walls, no heat is exchanged. Temperature increases from T_c to T_h. Work W_da is done.

The net work is |Q_h| – |Q_c|. For the entire cycle, ΔU = 0. The efficiency e = W/Q_h = 1 – Q_c/Q_h = 1 – T_c/T_h, which depends solely on temperatures.

Applications: Otto Engines & Passenger Cars

Cruise Control

Cruise control automatically maintains a steady speed in a motor vehicle. The cruising force (F) produced equals the resistance (R).

Energy Distribution in Conventional Cars

In a conventional car, fuel energy is distributed as follows:

Engine Losses: 70-72%
Parasitic Losses: 5-6%
Drivetrain Losses: 5-6%
Idle Losses: 3%
Power to Wheels: 17-21%

Improving Car Energy Efficiency

Converting more heat into kinetic energy using more efficient engines.
Reducing resistance (R) through improved car design.
Capturing energy wasted in braking (e.g., flywheel, regenerative braking).
Cruising at a slower speed.

Energy Consumption Calculation

Energy used per day = (Distance travelled / Distance per unit of fuel) × Energy per unit of fuel

Approximately 75% of energy from burning petrol is lost as heat. The remaining 25% is used for acceleration, overcoming air resistance (drag coefficient C_d), and overcoming road resistance (C_rr), with energy also dissipated during braking. Air resistance creates a tube of swirling air behind the car, moving at speed v.

Energy Dissipation in Cars

Braking Dissipation: Kinetic Energy (KE) / Time between braking = (1/2 m_c v²) / (d/v) = (1/2 m_c v³) / d, where m_c is the mass of the car.
Air Drag Dissipation:

Total Energy Dissipation

The rate of mechanical energy dissipated per second by the car = (1/2 m_c v³) / d + (1/2 ρ A v³)

If m_c/d > ρA (or m_c > ρAd, where ρAd is the mass of the tube of air), energy is primarily spent on braking. Otherwise, energy is mainly used to overcome air resistance.

d = m_c / (ρ A) = m_c / (ρ C_d A_c), where d is a special distance, approximately 750m.

For distances less than 750m, driving is dominated by braking. Beyond 750m, it is drag-dominated.

Otto Gasoline Engine Operation

The operation of a gasoline engine involves an intake stroke, a compression stroke, spark-initiated fuel combustion, a power stroke (where expanding gas does work), and a final exhaust stroke to expel residue gas.

Four-Stroke Otto Cycle

The petrol fuel injection internal combustion (IC) engine operates on a four-stroke Otto cycle, involving a piston moving within a hollow cylinder.

Fuel and air are first mixed in the carburetor.
Intake Stroke: Piston moves down, drawing fuel-air mixture into the cylinder.
Compression Stroke: Piston moves up, compressing the mixture into a small volume.
Power Stroke: Spark plug ignites the mixture. Expanding hot gas from combustion pushes the piston down, generating power for the car.
Exhaust Stroke: Piston moves up, ejecting burned gases from the cylinder.

IC engines typically have four or more cylinders, timed to ensure at least one cylinder is in the power stroke at any given moment.

Flywheel Energy Storage

Flywheel energy storage involves the engine providing motion to a disc/cylinder, increasing its speed during braking to store kinetic energy. This stored KE can later be drawn by the engine to accelerate the vehicle (reducing the flywheel’s rotational speed).

Electric Cars

Electric cars utilize batteries and electric motors, offering a clean and safe alternative. They provide faster acceleration but have shorter ranges and require longer charging times, though they produce no exhaust. Regenerative braking transfers kinetic energy from the wheels to the electric motor, which acts as a generator, sending electricity back to the battery.

Applications: Diesel Engines & Cargo Ships

Diesel vs. Petrol Engines: Key Differences

Petrol is generally more expensive, has lower energy content, and results in lower engine efficiency (20-30% for petrol vs. 50% for diesel). Petrol engines also have lower power output, and spark ignition can lead to violent engine knock due to pressure waves.

Diesel Engine Principles

A diesel engine is an internal combustion engine that ignites the air-fuel mixture using the heat of compressed air, rather than an electric spark. This results in high thermal efficiency.

Diesel engines dominate long-distance marine transport, powering oil tankers, container ships, and dry bulk carriers for grain, ores, and coal.

Four-Stroke Diesel Cycle

Intake: Air enters the combustion chamber.
Compression: All valves close, piston moves up, compressing air, increasing temperature and pressure.
Injection: Fuel is injected at high pressure.
Expansion: Fuel ignites upon contact with hot air, generating mechanical work.
Exhaust: Hot gases leave through the exhaust valve after combustion.

Diesel vs. Petrol: Performance Comparison

Diesel: Better energy efficiency, uses less fuel.
Petrol: Better cold start, less noise and vibration, more elasticity (better at high engine speeds), lighter, more power for the same engine size.

Engine Emissions

CO2 Emissions: Petrol engines have higher CO2 emissions per liter; Diesel engines have higher CO2 emissions per kilometer.
Diesel Emissions: Particulate Matter (PM), unburned hydrocarbons (HC), Carbon Monoxide (CO), Nitrogen Oxides (NOx).

Emission Control Technologies

Diesel:
- Exhaust Gas Recirculation (EGR) for NOx reduction.
- Particulate filters and oxidation catalytic converters for HC & CO.
- Selective Catalytic Reduction (SCR) converts NOx into N2 and H2O.
Petrol:
- Three-way catalytic converters perform oxidation (CO and HC into CO2 and H2O) and reduction (NO into N2 and O2).

Container Shipping Standards

Container boxes are typically made of corrugated weathering steel. Common sizes include Hi-cube (9 feet 6 inches high) and standard (8ft x 20ft x 8ft).

Other Prime Movers

Sparked Otto-cycle petrol engines are also considered prime movers.

Applications: Jet Engines & Air Travel

Forces in Flight

The four fundamental forces acting on an aircraft are thrust (forward force), weight, lift, and drag.

As an airplane moves at speed v through the air, the air flows over the aircraft and its wings in the opposite direction at the same speed.

Takeoff/Climb: Thrust > Drag and Lift > Weight
Cruise: Thrust = Drag and Lift = Weight
Landing: Thrust < Drag (increased by air brakes) and Lift < Weight, allowing the plane to slow down and descend.

Air Travel Energy Consumption

The estimated energy cost for air travel is approximately 30 kWh per person per day.

Power for Drag Overcome

Power required to overcome drag (P_drag) = (1/2) ρ C_d A_p v³, where C_d is the drag coefficient and A_p is the frontal area of the plane.

Downward Air Speed and Momentum

The mass of the air “sausage” is ρ v t A_s, where A_s is its cross-sectional area. Its momentum = mass × velocity = ρ v t A_s u. The change of momentum in time t, u = mg / (ρ v A_s).

Aircraft Fuel Efficiency

Energy per distance for a plane traveling at speed v, with an efficiency of approximately 1/3.

Thrust = (C_d A_p / A_s)^1/2 mg = (C_d f A)^1/2 mg

(Density in kg/m³)

Transport cost = (1/efficiency) × (thrust / mass), in N/kg.

Aircraft Range

Range is determined by a dimensionless factor multiplied by the fundamental fuel distance.

Bernoulli’s Principle

At a fixed height (h), the sum of pressure (p) and kinetic energy density is constant. Therefore, as speed (v) increases, pressure (p) decreases.

Important Note: For a thermodynamic process, if the gas expands, it does positive work; if compressed, it does negative work.

A force directed from high to low pressure areas develops whenever a pressure difference exists.

Jet Engine Components

A gas turbine (the core of a jet engine) has three major components: compressor, combustor, and turbine.

Homework Problems: Set 1

1. Bacterial Colony Growth

A bacterial colony starts growing in a jar at 11:00 am. The size of the colony doubles each minute, and the jar is just full at 12:00 noon.

(a) At what time was the jar 1/8 full? If the initial number of bacteria was 10. What was the number of bacteria when the jar was full?

(b) At what time was the jar (i) one thousandth, (ii) one millionth full?

Solution:

(a) The jar was half full at 11:59, quarter full at 11:58, and 1/8 full at 11:57. In one hour, the bacteria population doubles successively 60 times. The population after doubling 60 times = 10 × 2⁶⁰ = 10 × 2¹⁰ × 2¹⁰ × 2¹⁰ × 2¹⁰ × 2¹⁰ × 2¹⁰ ≈ 10 × (10³)⁶ = 10¹⁹

(b) Every minute back, the population is halved. One thousandth full is to go back 10 doublings, or 2¹⁰. The jar was one thousandth full at 11:50. One millionth full is to go back 20 doublings, or 20 minutes, i.e., at 11:40.

At 11:50, when the jar was just one thousandth full, or 99.9% empty, it was not conceivable that it would be full 10 minutes later. Three minutes later, at 11:53, it was 1/128 full. Another four minutes later, at 11:57, it was 1/8 full.

2⁰ + 2¹ + 2² + … + 2¹⁹ = 2²⁰ – 1 ≈ 10⁶

2⁰ + 2¹ + 2² + … + 2²⁹ = 2³⁰ – 1 ≈ 10⁹

2⁰ + 2¹ + 2² + … + 2⁵⁹ = 2⁶⁰ – 1 ≈ 10¹⁸

2⁰ + 2¹ + 2² + … + 2⁶³ = 2⁶⁴ – 1 ≈ 2⁴ × 2⁶⁰ = 16 × 10¹⁸

2. Compound Interest Calculations

The interest rate for time deposit in a bank in China is 3.5% per annum (the rate in Hong Kong is 0.4%). The interest rate for cash advances or retail purchases using credit card from a bank in Hong Kong is 35% per annum.

(a) If you make a 5-year deposit of $1000 in a bank in China, how much interest would you earn after 5 years?

(b) If you buy a $1,000 gift with your Hong Kong credit card and decide to pay the bank after 5 years. Find the interest you have to pay the bank.

(c) In the time period when your time deposit at a fixed rate of 3.5% is doubled, how much you have to pay the bank at a fixed rate of 35%?

Solution:

(a) The amount $A after 5 years at a rate of 3.5% is given by A = P(1 + 0.035)⁵ = 1000(1.035)⁵ ≈ 1000(1.19) = $1190. The interest = $190.

(b) The amount $A after 5 years at a rate of 35% is given by A = P(1+0.35)⁵ = 1000(1.35)⁵ ≈ 1000(4.48) = $4480. The interest = $3480.

(c) The doubling time at a rate of 3.5% is 20 years, while the doubling time at a rate of 35% is just 2 years. When your time deposit in a bank in China is doubled in 20 years, your payment to a Hong Kong bank would have doubled 10 times, i.e., 2¹⁰ ≈ 1000 times of your cash advance or retail purchase.

3. US Energy Consumption Analysis

In the United States, the total energy consumed per year per person is the equivalent of 58 barrels of oil. Express the energy consumption in kWh per person per day. (The energy in one barrel (41 gallons) of crude oil is equivalent to about 1700 kWh of heat.) For all the barrels of oil burned in the U.S., only 35.2 QBtu (quadrillion, or 10¹⁵, British thermal unit) per year are obtained as useful energy. Useful energy means mechanical or electrical energy. Express this useful energy in kWh per person per day. Take the population of the U.S. to be 293 million. How many equivalent barrels of oil per person per year are lost? (Hint: Find out from Google, 1 Btu = ? kWh). A heat engine can only convert a fraction of the heat energy into useful energy. What is the effective efficiency of the average equivalent heat engines used in the U.S.?

Solution:

The total energy equivalent in 58 barrels of crude oil = 58 × 1700 kWh. The equivalent energy use = 58 × 1700 / 365 = 270 kWh per person per day.

From Google, we find that 1 Btu = 0.000293 kWh. The useful energy per person per day = 35.2 × 10¹⁵ × 2.93 × 10^-4 kWh / (293 × 10⁶ × 365) = 96.4 kWh.

4. Earth’s Population Capacity

The Earth has a land area of 148,940,000 km². Assume that each human being requires a minimum of 10 m² of land to sustain her/his life.

(a) Estimate the Earth’s maximum possible population in billions. The present population is 7.0 billion people. If the human population grows at a rate of 2% per year, when will that maximum population be reached?

(b) Find the decrease in area (m²) per person when the maximum possible population is reached.

Solution:

(a) The maximum possible population = 148,940,000 × 1000² / 10 = 1.4894 × 10¹³ persons = 14,894 billion ≈ 7 × 2000 billion.

At a growth rate of 2%, in t years, the Earth’s population will be 7 × 10⁹ × e^0.02t. Equating this to 1.4894 × 10¹³, we have 0.02t = ln (14.894 / (7 × 10^-3)) ≈ 7.66.

(b) The present average area per person in the world is 20,000 m². This is 2,000 times 10 m². The area per person decreases by 2000 times.

(c) The doubling time for a 2% annual rate of increase is 35 years (rule of 70). In 383 years, there will be about 11 doublings. 2¹¹ ≈ 2000.

5. Weight Loss Exercise Calculation

(Losing weight the hard way) A student eats a Big Mac set meal at McDonald’s. The calories in a Big Mac burger, a Medium French Fries, and a Coke are respectively 560, 312, and 110 kcal. He wishes to do an equivalent amount of work in the gymnasium by lifting a 50.0 kg barbell. Assume he raises the barbell 2.0 m each time he lifts it and he regains no energy when he lowers the barbell. How many times must he raise the barbell to expend this much energy? If the student is in good shape and lifts the barbell once every 5 s, how long will it take him to finish his exercise? Take the mechanical efficiency of human muscles to be 25%.

Solution:

The total calories sum up to 982 kcal = 982 × 10³ × 4.186 J. The work done against gravity per lift = mgh = 50.0 kg × 9.8 m/s² × 2.0 m = 980 J. Since mechanical efficiency is 25%, the energy expended by the body per lift = 980 J / 0.25 = 3920 J.

Therefore, n = (982 × 10³ × 4.186 J) / 3920 J ≈ 1049 times. The time it takes to lift the barbell 1049 times = 1049 × 5 s = 5245 s ≈ 1.46 hours.

6. World Population Trends

(a) Find the increase in world population, from the graph in slide 18 of L01, between 1960 and 2000 and the estimated increase from 2000 to 2050.

(b) What can you say about the increase and rate of increase of world population before and after 1990?

(c) Compute and tabulate the increase and percentage increase of population in 1950, 1960, 1970, 1980, 2000, and the projected percentage increase in 2020, 2040.

Solution:

(a) The increase in population in 4 decades between 1960 and 2000 are respectively 680 million, 750 million, 830 million, and 790 million. The total increase = 680+750+830+790 = 3050 million, or about 3 billion. The population was doubled in 40 years. The increase in population in 5 decades between 2000 and 2050 are respectively 750+710+610+470+340 = 2880 million, less than 3 billion.

(b) Before 1990, the rate of population increase and the population increase every decade were generally rising. After 1990, the rate of population increase decreases, but the population keeps increasing, although at a slower pace.

7. Global Energy Supply

(a) In slide 14 of L03, the world’s total primary energy supply in 2012 was 13,371 Mtoe. Personalize this number to kWh per person per day, using the conversion table, slide 14 of L02 or from the internet. (b) How many 24-hour switched on 40 W light bulbs per day can 13,371 Mtoe of energy support?

Solution:

(a) The energy stored in 13,371 Mtoe = 13,371 × 10⁶ × 42 × 10⁹ J = (13,371 × 10⁶ × 42 × 10⁹ J) / (3.6 × 10⁶ J/kWh) kWh. The world population is 7 billion, thus the energy supply per person per day = (13,371 × 10⁶ × 42 × 10⁹) / (3.6 × 10⁶ × 7 × 10⁹ × 365) = 61 kWh.

(b) 1 kWh ≈ 24-hour switched on 40 W light bulb. Each person uses 61 bulbs, so 7 billion people switch on 427 billion bulbs.

Homework Problems: Set 2

1. Isothermal Expansion

1.0 mol sample of an ideal gas is kept at 0.0 °C during an expansion from 3.0 liter to 10.0 liter.

(a) How much work is done on the gas during the expansion?

(b) How much energy transfer by heat occurs with the surroundings in this process?

(c) If the gas is returned to the original volume by means of a constant pressure process, how much work is done on the gas?

Solution:

(a) This is an isothermal expansion, W = nRT ln(V_i/V_f) = 2.7 × 10³ J.

(b) There is no change of temperature, so the change of internal energy is zero, ΔE = Q + W = 0, Q = 2.7 × 10³ J.

2. Carnot Cycle and Efficiency

(a) Consider the Carnot cycle as shown in the p-V diagram. Find the entropy change at T_H and at T_C. Hence, construct the T-S diagram of the Carnot cycle.

(b) Treating the human body (37°C) as a heat engine, what is its possible maximum efficiency if the room temperature is 20°C?

Solution:

(a) For the expansion from A to B at constant temperature T_H, the entropy change is ΔS_h = Q_h/T_H. The contraction from C to D takes place at constant temperature T_C, the entropy change is ΔS_c = Q_c/T_C. The entropy change is zero in the adiabatic expansion BC and contraction DA because there is no heat enters or leaves the system.

(b) Maximum efficiency is the Carnot efficiency = 1 – T_c/T_h. T_h = 310 K, T_c = 293 K. Max efficiency = 17/310 = 5.5%.

3. Carnot Engine Performance

Imagine a Carnot engine that operates between temperatures T_H = 850 K and T_L = 300 K. The engine performs 1200 J of work each cycle, which takes 0.25 s.

(a) What is the efficiency of the engine?

(b) What is the average power of this engine?

Solution:

(a) e = 1 – T_L / T_H ≈ 0.647 ≈ 65%.

(b) Power = work/time = 1200 J / 0.25 s = 4800 J/s = 4.8 kW.

4. Adiabatic Compression in Engines

Air is always compressed in an internal combustion engine approximately adiabatically. The compression ratio is typically about 8 in a petrol engine, between 15 and 25 in a diesel engine, and up to 44 in a gas turbine. Find the temperature of air after compression for the compression ratio mentioned above, assuming the adiabatic process is ideal and the initial temperature is 300K.

Solution:

For adiabatic process, we have PV^γ = constant. P = nRT / V, thus TV^γ-1 = constant. The temperature after adiabatic compression is given by T_f = T_i (V_i / V_f)^γ-1.

5. Car Power and Energy Consumption

Find the power needed to keep a car traveling at a speed 100 km/h. If the speed is increased to 125 km/h, what is the power needed to keep the car moving? By what factor is the power used increased or decreased?

Find the energy used in covering 100 km with speed of 100 km/h and 125 km/h, using the data given by MacKay. Find the factor the energy used is increased or decreased.

Solution:

The power needed to keep a car moving at a speed v = 1/2 ρAv³ watt.

ρ = 1.3 kg/m³, A = 1 m². 100 km/h ≈ 27.8 m/s, power moving at 100 km/h ≈ 14 kW. 125 km/h ≈ 34.7 m/s, power moving at 125 km/h ≈ 27.2 kW.

The power needed is increased by a factor of 27.2/14 ≈ 1.94 = (125/100)³. Increasing the speed by 25% almost doubles the power needed.

Note: power/speed = energy/distance.

The energy per m at a speed of 100 km/h = 503.6 J (here we take the ratio of 14kW/27.8m/s). The energy to cover 100 km = 503.6 × 10⁵ J = 14 kWh.

The energy per m at a speed of 125 km/h = 782.7 J. The energy to cover 100 km = 782.7 × 10⁵ J = 21.7 kWh.

The energy used is increased by a factor of 21.7/14 ≈ 1.56 = (125/100)².

Alternatively, the energy to cover 100 km can be obtained by multiplying the hours needed to cover the distance. The time it takes to cover 100 km at 100 km/h is 1 hr. The time it takes to cover 100 km at 125 km/h is (100/125) = 0.8 h.

Energy per 100 km travelled moving at 100 km/h = 503.6 × 10⁵ J = 14 kWh.

Energy per 100 km travelled moving at 125 km/h = 27.2 × 0.8 = 21.7 kWh.

Homework Problems: Set 3

1. Container Ship Energy Consumption

Consider the containership, Ever Uberty cited by MacKay (p.95), length 285 m, width 40 m, a service speed of 25 knots (46.3 km/h) is powered by a 44 MW engine. The engine efficiency is 50%. It has a container capacity of 4948 TEU. One TEU is the size of a small 20-foot container with a volume of about 40 m³. Most containers in use are 40-foot containers with a size of 2 TEU weighing 4 tons and a freight capacity of 26 tons. Show that this ship’s energy consumption for cargo is 0.03 kWh of chemical energy per ton-km.

Solution:

Note (energy/mass × distance) = (power/mass × speed).

Power = 44 MW, speed = 46.3 km/h.

The weight (actually mass) of cargo, m = (4948/2) × 26 t = 64,324 t (assuming 40-foot containers are 2 TEU, so 4948 TEU means 2474 40-foot containers).

The 44 MW power (P) of the container ship is derived from the chemical energy of the fuel. (But only half of the chemical energy is used to propel the ship).

The chemical energy cost of transport of the cargo = 2 × (power / (mass × speed)) MW / (t-km/h) = 2 × (power / (mass × speed)) × 10³ kWh / (t-km)

= 2 × 44 × 10³ / (64324 × 46.3) = 2 × 0.0148 ≈ 2 × 0.015 = 0.03 kWh/t-km.

2. Oil Tanker Energy Efficiency

MacKay cited a modern oil tanker (p.95) uses 0.017 kWh/t-km, with a cargo weight of 40,000 t, capacity of 47,000 m³. The tanker is powered by engine with maximum output of 11.2 MW. Its cruise speed at 8.2 MW is 15.5 knots (29 km/h). How much energy is carried in a fully loaded ship? Find the energy needed to ship the oil cargo one-quarter of the way around the world (10,000 km) as a percentage of the energy the ship is carrying.

Solution:

The oil tanker uses energy at the rate of 0.017 kWh/t-km. The energy used to transport 40,000 t of cargo 10,000 km = 0.017 × 40,000 × 10,000 = 6.8 × 10⁶ kWh.

47,000 m³ = 47,000 × 1000 liters = 47 × 10⁶ liters. The calorific value of one liter of crude oil = 37 MJ (Sustainable Energy, p.334). The chemical energy in 47,000 m³ of crude oil = 47 × 10⁶ × 37 MJ = 1.739 × 10⁶ GJ = 1.739 × 277.8 × 10⁶ kWh = 483 × 10⁶ kWh.

The energy cost of the crude oil by the oil tanker is (6.8 / 483) × 100 % = 1.4% of the crude oil cargo.

3. Supertanker Stopping Distance

Large tankers have a large inertia, making them very difficult to steer. A loaded supertanker could take as much 10 km and 15 minutes to come to a full stop. This can be shown in the case for the ULCC TI Europe. The power output of the diesel engine of the TI Europe is 37 MW, its cruising speed is 16.5 knots.

(a) Find the resistive force the tanker has to overcome moving at cruising speed.

(b) Find the stopping distance, the distance travelled after the engine is turned off.

The mass of the loaded tanker = 440,000 × 1000 kg = 440 × 10⁶ kg.

The power output of the engine = 37 MW.

The cruising speed of the tanker = 16.5 knots = 8.5 m/s.

Solution:

(a) Overcoming the resistive force F to keep the tanker moving with speed v requires the engine to provide mechanical power equal to Fv.

We have 37 × 10⁶ = F (8.5), F = 4.35 × 10⁶ N.

(b) The kinetic energy of the tanker is dissipated as work done against F covering the stopping distance D. The kinetic energy = (1/2) × (440 × 10⁶) × (8.5)² = (4.35 × 10⁶) D.

D = (220 × (8.5)²) / 4.35 ≈ 3.7 km.

(c) To find the stopping time, we use 0 = v – at. The deceleration a is given by a = F/m = 4.35 × 10⁶ / (440 × 10⁶) m/s².

t = v/a = (8.5 / (4.35/440)) ≈ 860 s ≈ 14.3 minutes.

4. Boeing 747-400 Range Calculation

The empty weight (mass), the maximum take off weight (MTOW, mass), the maximum fuel capacity of a Boeing 747-400 aeroplane and the maximum range at MTOW obtained from Boeing are 179t, 397t, 216,840L and 13,450 km, respectively. For a typical seating of 524 passenger, if each passenger carries half of her/his weight of baggage, what is range of the plane if in addition, it also carries 10 tonnes of cargo? Assume the mass of an average passenger is 80 kg. The specific density of fuel (kerosene) is 0.817. The weight of 1 m³ of water is 1 tonne. Assume that the range is proportional to the fuel the plane carries.

Solution:

For a Boeing 747-400, maximum take-off weight (MTOW) = 397 t, empty weight = 179 t, maximum fuel capacity = 216,840 L, maximum range at MTOW = 13,450 km.

The volume of fuel at maximum capacity (for maximum range) = 216.8 m³. The corresponding weight = 216.8 × 0.817 = 177 t.

The weight of 524 passengers and baggage (80 kg passenger + 40 kg baggage = 120 kg/person) = 524 × 120 kg = 62880 kg ≈ 63 t.

The weight of passengers, baggage, and cargo = 63 t + 10 t = 73 t.

The total weight of the aircraft, passenger, baggage, and cargo = 179 t (empty) + 73 t (payload) = 252 t.

The maximum fuel that it can carry under the present condition = 397 t (MTOW) – 252 t (aircraft + payload) = 145 t.

The fraction of maximum fuel capacity = 145 / 177 ≈ 81.9%.

The corresponding range = 13,450 km × 0.819 ≈ 11,016 km.

5. Air vs. Road Fatality Rates

The world airline fatalities in 2005 and 2010 were 1050 and 817 respectively. The passenger-km in 2005 and 2010 were about 3,700,000 million and 4,700,000 million respectively, find the fatality rate per 100 million passenger-km in 2005 and 2010. Giving the US road fatality rate per 100 million VMT is about 2, the average passenger-car occupancy is 1.5 and the average yearly fatality of USA to be 40,000, how much is the rate you have obtained compared with the road traffic fatality rate of the USA?

Solution:

The passenger-km in 2005 and 2010 were about 3,700,000 million and 4,700,000 million respectively.

The fatality rate per 100 million passenger-km in 2005 = 1050 / (3,700,000 × 10⁶ / 100 × 10⁶) = 1050 / 37,000 = 0.028.

The fatality rate per 100 million passenger-km in 2010 = 817 / (4,700,000 × 10⁶ / 100 × 10⁶) = 817 / 47,000 = 0.017.

The US road fatality rate per 100 million VMT is about 2. One mile is about 1.6 km. If the average passenger car occupancy is 1.5, the fatality rate per 100 million passenger-km is about 2 / 1.6 / 1.5 ≈ 0.83. This is about 0.83 / 0.028 ≈ 30 times worse than air fatality in 2005, and 0.83 / 0.017 ≈ 49 times worse than air fatality in 2010. In terms of total deaths, it is about 40,000 / (1050/3700000*10^6 * (total US passenger-km)) which is not directly comparable without total US passenger-km for road. However, comparing the rates directly, road travel is significantly more fatal per passenger-km.

6. Aeroplane Transport Cost

MacKay has shown that the transport cost of an aeroplane, given as force divided by the product of engine efficiency and mass of the plane (C-24, p. 274), is 0.15 g (g is the acceleration due to gravity). Show that this is the same as 0.4 kWh/ton-km. Taking the cargo weight to be one fourth of the weight of the 747 plane, find the cargo transport cost of the 747 per ton-km.

Solution:

Transport cost is energy used per unit mass per distance travelled which is effectively the force per unit mass = 0.15 g N/kg = 0.15 × 9.81 N/kg ≈ 1.47 N·m/kg·m = 1.47 J/kg·m.

To convert J/kg·m to kWh/ton-km:

1.47 J/kg·m = 1.47 × (1/3.6 × 10⁶) kWh / (1/1000) ton × (1/1000) km = 1.47 × 10⁶ / 3.6 × 10⁶ kWh/ton-km ≈ 0.408 kWh/ton-km ≈ 0.4 kWh/ton-km.

The cargo weight is 1/4 of the total weight, so the transport cost of cargo per ton-km should be increased by a factor of 4, which is 4 × 0.4 kWh/ton-km = 1.6 kWh/ton-km.